Question

The frequency factor and activation energy for a chemical reaction are A = 4.23 x 10–12 cm3/(molecule·s) and Ea = 12.9 kJ/mol at 384.7 K, respectively. Determine the rate constant for this reaction at 384.7 K.

Answer 1

Answer : The rate constant for this reaction at 384.7 K is,
`7.493* 10^(-14)cm^3mole^(-1)`

Explanation :

The relation between frequency factor, rate constant and activation energy for a chemical reaction is,

`k=A* e^{[(-Ea)/(RT)]}`

where,

k = rate constant = ?

A = frequency factor =
`4.23* 10^(-12)cm^3\text{ molecule}^(-1)s^(-1)`

Ea = activation energy = 12.9 kJ/mol

T = temperature = 384.7 K

Now put all the given values in this formula, we get:

`k=4.23* 10^(-12)cm^3\text{ molecule}^(-1)s^(-1)* e^{[(-12.9kJ/mol)/((8.314J/mole.K)* (384.7K))]}`

`k=7.493* 10^(-14)cm^3mole^(-1)`

Therefore, the rate constant for this reaction at 384.7 K is,
`7.493* 10^(-14)cm^3mole^(-1)`